Question

the perimeter of rectangle is 40cm. the length of the rectangle is more than double its breadth by 2. find the length and breadth

Answer 1

P=2l+2w

We are given, 

P=40cm

l=2w+2

Plug these into the first formula and solve for the dimensions,

length and width, of the rectangle:

 

40=2(2w+2)+2w              Simplify

40=4w+4+2w

40=6w+4

36=6w

6=w OR the width is 6cm, therefore, 

the length is

l=2(6)+2

l=14cm

 

To calculate the area of a rectangle, A=lw,

plug the length,14cm, and the width, 6cm, into this formula

A=(14)(6)

A=84cm2 OR the area of the rectangle is 84cm2.

Answer 2

$\bf{\underline{\underline \blue{Solution:-}}}$

$\sf\underline{\red{\:\:\: AnswEr:-\:\:\:}}$

The breadth of the rectangle = 6 Cm

The length of the rectangle = 14 Cm

$\sf\underline{\red{\:\:\: Given:-\:\:\:}}$

The perimeter of rectangle is 40cm.

The length of the rectangle is more than double its breadth by 2

$\sf\underline{\red{\:\:\: Need\:To\: Find:-\:\:\:}}$

The breadth of the rectangle = ?

The length of the rectangle = ?

$\bf{\underline{\underline \blue{Explanation:-}}}$

$\sf\underline{\pink{\:\:\: Diagram:-\:\:\:}}$

$\setlength{\unitlength}{2cm}\begin{picture}(16,4)\thicklines\put(8,3){\circle*{0.1}}\put(7.8,3){\large{D}}\put(7.2,2){\mathsf{\large{?cm}}}\put(8,1){\circle*{0.1}}\put(7.8,1){\large{A}}\put(9.3,0.8){\mathsf{\large{?cm}}}\put(11.1,1){\large{B}}\put(8,1){\line(1,0){3}}\put(11,1,){\circle*{0.1}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11,3){\circle*{0.1}}\put(11.1,3){\large{C}}\end{picture}$

Let the breadth of the rectangle = y

$\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}$

Length = 2y + 2

$\sf\underline{\red{\:\:\: Formula\:Used\: Here:-\:\:\:}}$

$\bigstar \: \boxed{ \sf \: Perimeter\:of\:a\: rectangle = 2 \times (Length + Breadth) }$

$\sf\underline{\green{\:\:\: Now,Putting\:the\: values:-\:\:\:}}$

$\dashrightarrow \sf {Perimeter \:of\:a\: rectangle = 2[(2y + 2) + (y)]}$

$\dashrightarrow \sf {40 = 2 \times (3y + 2) }$

$\dashrightarrow \sf {40 = 6y + 4}$

$\dashrightarrow \sf {40 - 4 = 60}$

$\dashrightarrow \sf {y = \dfrac{\cancel{36}}{\cancel{6}}\:}$

$\dashrightarrow \sf {y = 6}$

$\sf\underline{\green{\:\:\: ThereFore:-\:\:\:}}$

The breadth of the rectangle = y

The breadth of the rectangle = 6 Cm

$\sf\underline{\green{\:\:\: And:-\:\:\:}}$

The length of the rectangle = 2y + 2

The length of the rectangle = 2 × 6 + 2

The length of the rectangle = 12 + 2

The length of the rectangle = 14 Cm

$\rule{200}{2}$